Partial sums of the cotangent function

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Partial sums of the cotangent function

Sandro Bettin, Sary Drappeau

To cite this version:

Sandro Bettin, Sary Drappeau. Partial sums of the cotangent function. Journal de Théorie des Nom- bres de Bordeaux, Société Arithmétique de Bordeaux, 2020, 32 (1), pp.217-230. �10.5802/jtnb.1119�.

�hal-02189191�

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S. BETTIN AND S. DRAPPEAU

Abstract. Nous prouvons l’existence de formules de r´eciprocit´e pour des sommes de la formePk−1

m=1f(^m_k) cot(π^mh_k ), oùfest une fonctionC¹par morceaux, qui met en évidence un phénomène d’alternance qui n’apparaˆıt pas dans le cas classique oùf(x) =x. Nous déduisons des majorations de ces sommes en termes du développement en fraction continue deh/k.

We prove the existence of reciprocity formulae for sums of the formPk−1 m=1f ^m_k

cot π^mh_k wheref is a piecewise C¹ function, featuring an alternating phenomenon not visible in the classical case wheref(x) =x. We deduce bounds for these sums in terms of the continued fraction expansion ofh/k.

1. introduction

There are several results in the literature proving reciprocity formulae for certain averages of the cotangent function. A prototypical example is the classical Dedekind sum which can be defined as

s h

k

:=− 1 4k

k−1

X

m=1

cot πm

k

cot

πmh k

, h, k ∈N, (h, k) = 1, and satisfies the well known reciprocity formula

s h

k

+s k

h

− 1

12hk = 1 12

h k +k

h −3

.

The Dedekind function has been generalized in several ways, all satisfying some sort of reciprocity (see e.g. [Zag73, Bec03, Ber76]).

Another related example is given by the Vasyunin sum V

h k

:=

k−1

X

m=1

m k cot

πmh

k

, (h, k ∈N, (h, k) = 1),

where, here and in the following, the overline indicates any multiplicative inverse modulo the denominator. The Vasyunin sum satisfies the reciprocity formula

(1.1) V

h k

+V

k h

= log 2π−γ

π (k+h) +k−h π logh

k−

√ hk π²

Z ∞

−∞

ζ ¹₂ +it

2 h k

it

dt

1 4 +t², whereζ(s) is the Riemann zeta-function, as well as another one relatingV h/k

withV k/h . For this, other generalizations, and the relation of V with the B´aez-Duarte and Nyman- Beurling criterion for the Riemann hypothesis we refer to [Vas95, BC13a, BC13b, BD03, Bag06].

2010Mathematics Subject Classification. 11L03 (primary); 11A55, 11M35 (secondary).

Key words and phrases. cotangent sum, continued fraction.

1

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Following the Euclid algorithm and repeatedly applying these results, one can then obtain bounds and asymptotic formulae for these sums in terms of the continued fraction expansion ofh/k. See, for example [Hic77] and [Bet15].

All of the results mentioned above involve sums over all (non-zero) residues. Having in mind an application to the value distribution of Kashaev’s knot invariants [BD], we are led to the problem of bounding partial sums of the cotangent function. For example, let

C_` h

k

:= 1 k

X

1≤m≤`

cot

πmh k

, h∈Z, k∈N, (h, k) = 1, 1≤` < k.

Despite the simplicity of this definition, this partial average behave rather erratically, and there is little reason to suspect at first the existence of pure reciprocity formulae such as (1.1).

0.2 0.4 0.6 0.8 1.0

-150 -100 -50 50 100 150

0.2 0.4 0.6 0.8 1.0

-40 -20 20 40

Figure 1. The graphs of (₆₇₇^` , C_`(²³¹₆₇₇)) as 1≤` <677 and of (₂₁₅^` , C_`(₂₁₅¹⁶) as 1≤` <215.

Figure 1 shows the evolution of C_`(231/677) and C_`(16/215) as ` varies. The continued fraction expansions of ²³¹₆₇₇ and ₂₁₅¹⁶ are {0; 2,1,13,2,3,2} and {0; 13,2,3,2} respectively, so that the similarity of the two graphs suggests that a reciprocity formula is in action. As we will see below, in this case, the reciprocity formula which we naturally obtain doesn’t directly relateC_`(h/k) withC_`(k/h), nor C_`(h/k) withC_` k/h

, but rather C_` h/k

withC_`⁰(h/k₁), wherek1 ≡k1 (modh) and 0< k1 ≤h, that is, the double iteration of a standard reciprocity formula. This seems to be the first observed instance of such an alternating behaviour in these objects. We give this (non-exact) reciprocity formula in Corollary 4 below.

In general we will consider this question for S_f

h k

:= 1

k

k−1

X

m=1

fm k

cot

πmh

k

, h∈Z, k∈N, (h, k) = 1.

wheref is a piecewiseC¹ function. It will follow from our arguments that for generic functions f,S_f ^h_k

doesn’t satisfy a simple reciprocity formula unlessf is smooth with only one point of discontinuity other than, possibly, 0. However, we are able to control its size in all cases, as needed for example in our application to the distribution of the Kashaev invariants.

Theorem 1. Let f : R→ R be a 1-periodic function which is piecewise C¹, and continuous except possibly at d points. Let D₀ = maxx∈Rlimε→0|f(x+ε)−f(x−ε)| and D₁ := kf⁰k₂.

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Then

Sf

h k

≤ dD₀ πk

r

X

m=1

vmlog(_v_m−1^v^m ) +O(dD0+D1),

Sf

h k

≤ dD0

π

r−1

X

m=0

log(^v^m+1_v

m )

v_m +O(dD0+D1),

where v₀, . . . , v_m are the partial quotients of the continued fraction expansion of h/k.

Remark 1. Following the approach of [Bet15], it would be possible to compute the distribution ofS_f ^h_k

as 1≤h < kandk→ ∞(the method of [MR16] might also be used with some effort).

In particular, one can prove that there is a functionε :R>0 → R>0 with limA→∞ε(A) = 0, so that for each A > 0, the number of h ∈ [1, k] with (h, k) = 1 and

Sf h k

≤ A is at leastk(1−ε(A)) +o(k) as k→ ∞.

Remark 2. If{x}denotes the fractional part of x, thenπcot(πx)− {x}⁻¹− {−x}⁻¹ extends to an odd bounded function onR. In particular, Theorem 1 holds also if one replaces cot(πy) by ({y}⁻¹− {y}⁻¹)/π in the definition ofS_f ^h_k

(or any other 1-periodic function ofy having a similar asymptotic behaviour around 0).

Theorem 1 is a crucial ingredient in the following law of large numbers for values of the Kashaev invariants of the figure eight knot, which we prove in [BD] :

Theorem 2. Forx∈Q, let

J(x) :=

∞

X

n=0 n

Y

r=1

1−e^2πirx

2

be the Kashaev invariant of the 4₁ knot. For some constantµ >0, as Q→+∞, we have logJ(x)∼µ(logQ) log logQ

for a proportion 1−o(1)of fractions x∈(0,1]∩Q of reduced denominator at most Q.

2. A reciprocity formula

2.1. Lemmas. We introduce some notation and give some basic results on the Hurwitz and periodic zeta-functions (cf. [Apo76, Chapter 12.9]). Forx∈R and<(s)>1, let

ζ(s, x) := X

n+x>0

1

(n+x)^s, F(s, x) :=X

n≥1

e(nx) n^s ,

where e(x) := e^2πix. Notice that we have “periodized” the Hurwitz zeta-function. It is well known thatζ(s, x) and F(s, x) extend as meromorphic functions on C and satisfy functional equations. The functional equations are nicely expressed in terms of ζ^±(s, x) := ζ(s, x) ± ζ(s,−x) and F^±(s, x) =F(s, x)±F(s,−x). Indeed, they become

F⁺(s, x) =χ(s)ζ⁺(1−s, x), F⁻(s, x) = 2iΓ(1−s) (2π)^1−s cos

πs 2

ζ⁻(1−s, x).

whereχ(s) := 2_(2π)^Γ(1−s)1−s sin(^πs₂ ) is as in the functional equation of the Riemann zeta-function.

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We also recall the special values

(2.1)

F⁺(1, x) =−log((1−e^−2πix)(1−e^2πix)) =−log(4 sin(πx)²), x /∈Z, F⁻(1, x) = 2πiX

n∈N

sin(2πnx)

πn = 2πi((x)) where ((x)) := 0 ifx∈Zand ((x)) := ¹₂ − {x} otherwise.

Also, we have the expansion ats= 1 ζ(s, x) = 1

s−1 −ψ({x}) +O(s−1), whereψ is the digamma function [GR07,§8.36], and so also (2.2) F⁺(1−s, x) =−1− γ+ log 2π+¹₂ψ({x}) +¹₂ψ({−x})

(1−s) +O((s−1)²).

Forh, `∈Z,k∈N we define (2.3) V₁

h k,`

k

:= 2X

m≥1

X

n≥1

sin(2πnm^h_k) cos(2πm_k^`)

πnm = 2X

m≥1

cos(2πm_k^`) m

mh k

which clearly converges sincePk

m=1cos(2πm`/k)((mh/k)) = 0 by parity. Actually, the above computation together with the functional equations forζ(s, x) can be used to show that one can smoothly truncate the sum overm, n at height Xk^1+ε, for anyX ≥1, at the cost of an error which isO((Xk)⁻¹⁰⁰), as is done e.g. in [Bet15, p. 11423] with an analogous series.

Forh, p∈Z,k, q∈N, (h, k) = 1 we also define V₂(^h_k,^p_q) := X

m∈Z

|m+p q|≥1

X

n≥1

sin(2π^h_kn|m+ ^p_q|)

πn|m+^p_q| = X

m∈Z

|m+p q|≥1

h

k|m+^p_q|

|m+^p_q|

where the outer sum is computed by summing together the terms m and −m. Again, one easily sees that the series converges and that one can smoothly truncate the sums atm, n X(qk)^1+ε at a cost of an error which is O((Xqk)⁻¹⁰⁰). The manipulation of conditionally convergent sums and integrals in the following is justified by these considerations.

We remark (cf. e.g. Lemma 3 below with f(x) = x for 0 < x < 1) that V₁(^h_k,_k^`) and V2(^h_k,^p_q) reduce to −^π_kV(^h_k) if k|` and p= 0.

We will prove a reciprocity formula relating V1 with V2, generalizing (1.1), following the same approach as in the proof of [BC13a, Theorem 5]. We shall need the following uniform convexity bound forF⁺(¹₂+it, x).

Lemma 1. Let x6∈Z and

K⁺(s, x) :=F⁺(s, x)−χ(s)({x}^s−1+{−x}^s−1).

ThenK⁺(¹₂ +it, x)_ε(1 +|t|)¹⁴^+ε uniformly in x.

Proof. For <(s) = 1 +ε we have K⁺(s, x) _ε 1, whereas on <(s) = −ε after applying the functional equation and expanding the resulting series we haveK⁺(s, x) _ε (1 +|s|)¹²^+ε uniformly inx. Phragmen-Lindel¨of’s theorem [Tit39,§5.61] then gives the claimed bound.

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2.2. Main reciprocity formula.

Proposition 1. Let `, h, k ≥ 1 and (h, k) = 1. For h 6= 1, let β := {^k_h}⁻¹{_h^`} if k - ` and β= 0 if k|`. Then, we have

(2.4) 1 hV1

h k,`

k

+δh6=1

1 kV2

nk h

o , β

=

γ_h,k,`

h − 1 k

log k

h

+O 1

k+ 1 h

,

where δh6=1 = 0 if h = 1 and δh6=1 = 1 otherwise, and where γh,k,` = 1 if k|` and otherwise γ_h,k,`= 2 if k≤h and 0≤γ_h,k,`≤1 if k > h.

Remark 3. Fork-`, the termγ_h,k,`log ^k_h

/h can be replaced in this estimate by

1

h log{_k^`}+ log{−^`_k} −log⁻({^`_k}^k_h)−log⁻({−^`_k}^k_h) , where log⁻(x) := min(logx,0).

Proof. We assumek-`, since otherwise the formula is an immediate consequence of the usual Vasyunin’s formula (1.1). Let

I(h, k) := 1 2πi

Z

(¹₂)

F⁺(s,^`_k)ζ(1−s) h^sk^1−s

ds s(1−s), whereR

(c)·ds:=Rc+i∞

c−i∞ ·ds. We will now proceed to evaluateI(h, k) in two different ways.

Throughout, we denote

α= ` k. In the notation of Lemma 1, we have

I(h, k) = 1 2πi

Z

(¹₂)

K⁺(s, α)ζ(1−s) h^sk^1−s

ds

s(1−s)+ 1 2πi

Z

(¹₂)

χ(s)({α}^s−1+{−α}^s−1)ζ(1−s) h^sk^1−s

ds s(1−s). By Lemma 1, the first integral is bounded by

1

√ hk

Z

R

|ζ(¹₂+it)|dt (1 +|t|)³²

1

√ hk, whereas the contribution of{α} to the second integral is

1 2πi

Z

(¹₂)

χ(s){α}^s−1ζ(1−s) h^sk^1−s

ds

s(1−s) = 1 k{α}

1 2πi

Z

(¹₂)

ζ(s)

{α}k h

s

ds s(1−s).

If{α}^k_h <1 we move the line of integration to<(s) = +∞ obtaining a contribution from the double pole ats= 1 of

log({α}^k_h) +γ−1

h .

If{α}^k_h ≥1 then we move the integral to<(s) =−¹₄ passing through a simple pole ats= 0.

The contribution of the residue is −_2k{α}¹ = O(1/h), whereas the integral on the new line contributesO(_k{α}¹ {α}^k_h−1/4

) =O(1/h) sinceζ(−1/4 +it)(1 +|t|)³⁴. Thus, in both cases we find that the contribution of{α}to the second integral is

log⁻({α}k/h) +O(1)

h .

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Repeating the same computation for{−α}we then obtain our first expression for I(h, k):

(2.5) I(h, k) = log⁻({α}k/h) + log⁻({−α}k/h) +O(1)

h +O

1

√ kh

.

Now we computeI(h, k) in a second way. We split the integral into 1

2πi Z

(¹₂)

F⁺(s, α)ζ(1−s) h^sk^1−s

ds s(1−s)

= 1 2πi

Z

(¹₂)

F⁺(s, α)ζ(1−s) h^sk^1−s

ds 1−s+ 1

2πi Z

(¹₂)

F⁺(1−s, α)ζ(s) h^1−sk^s

ds

1−s =I₁+I₂, say, where in the second integral we made the change of variabless→1−s. We now consider I1. We move the line of integration to<(s) = 5/4 passing through a simple pole ats= 1. We obtain

I₁ =−F⁺(1, α)

2h + 1

2πi Z

(5/4)

F⁺(s, α)ζ(1−s) h^sk^1−s

ds 1−s. By (2.1),

F⁺(1, α) =−log(4 sin(πα)²) =−2 log{α} −2 log{−α}+O(1).

Also, 1 2πi

Z

(5/4)

F⁺(s, α)ζ(1−s) h^sk^1−s

ds

1−s = X

m∈Z6=0

X

n≥1

e(mα) k

1 2πi

Z

(5/4)

χ(1−s)(hn|m|/k)^−s ds 1−s. By the Mellin formula

1 2πi

Z

(5/4)

χ(1−s)u^−s ds

1−s = sin(2πu) πu we find

1 2πi

Z

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F⁺(s, α)ζ(1−s) h^sk^1−s

ds

1−s = X

n≥1,m∈Z6=0

sin(2πhn|m|/k) e(mα)

πhn|m| = 1

hV₁ h

k, α

.

Thus,

I₁ = log{α}+ log{−α}+O(1)

h + 1

hV₁ h

k, α

.

As forI₂we move the line integration to<(s) = 5/4 passing through a double pole ats= 1.

By (2.2) the pole contributes a residue (2.6) ψ({α}) +ψ({−α}) + 2 log(2πk/h)

2k = −{α}⁻¹− {−α}⁻¹+ 2 log(k/h) +O(1) 2k

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sinceψ(x) = ¹_x+O(1) for 0< x <1. The contribution of the integral toI₂ is 1

2πi Z

(5/4)

F⁺(1−s, α)ζ(s) h^1−sk^s

ds

1−s = X

m∈Z

X

n≥1

1 2πih

Z

(5/4)

χ(1−s) kn

h |m+α|

−s

ds 1−s

= X

m∈Z

X

n≥1

sin(2π^k_hn|m+α|) πkn|m+α|

= 1 k

X

m∈Z k

h|m+α|

|m+α| ,

where in the first line we have applied the functional equation to F⁺ and expanded the Dirichlet series.

Now, we observe that the series sums to zero if h = 1 (since k|m+α| ∈ Z in this case) and the claimed result easily follows. Thus, assume h 6= 1. We isolate the terms m ∈ {−bαc,−bαc −1,b−βc,b−βc+ 1} from the sum, where

β:=

nk h

o−1n` h

o . The termsm∈ {−bαc,−bαc −1}contribute

k h{α}

k{α} +

k

h{−α}

k{−α}

whereas the contribution of m⁰ ∈ {b−βc,b−βc+ 1} (with m⁰ 6=−bαc,−bαc −1) is bounded by

1 k

k

h|m⁰+α|

|m⁰+α| 1 k.

Next, we replaceα in the denominator byβ. The error in doing so is 1

k

X

m6=−bαc,−bαc−1,b−βc,b−βc+1 k

h|m+α|

1

|m+{α}|− 1

|m+β|

1

k.

We can then include again the terms −bαc and −bαc − 1 (when they are different from b−βc,b−βc+ 1) at the cost of an error which is analogously seen to be O(1/k), obtaining

1 k

X

m∈Z k

h|m+α|

|m+α| = 1 k

X

m∈Z, m6=b−βc,b−βc+1

k

h|m+α|

|m+β| +

k h{α}

k{α} +

k

h{−α}

k{−α} +O 1

k

.

By periodicity of the sine we then have 1

k

X

k

h|m+α|

|m+β| = 1 k

X

k

h|m+β|

|m+β| = 1 kV2

nk h

o , β

+O

1 k

.

Finally, we observe that

k h{α}

k{α} +

k

h{−α}

k{−α} = 1

2k{α} + 1

2k{−α}−{_h^k{α}}

k{α} −{^k_h{−α}}

k{−α} +O 1

h

= 1

2k{α} + 1

2k{−α}+O 1

h

.

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Indeed, the third term is O(1/h) (and similarly for the fourth) since, if 0 ≤ `⁰ ≤ k and 0 ≤

`⁰⁰≤h are such that `⁰ =` (modk) and `⁰⁰ ≡`⁰ (modh), then {_h^k{α}}/k{α}=`⁰⁰/(`⁰h)≤ _h¹. By the above computations we then have

1 2πi

Z

(5/4)

F⁺(1−s, α)ζ(s) h^1−sk^s

ds 1−s = 1

kV₂({^k_h}, β) + 1

2k{α} + 1

2k{−α} +O 1

k +1 h

and so, adding the contribution of the residue (2.6) we have I₂=V₂({_h^k}, β)/k+ log(k/h)/k+O

1 k +1

h

. Thus,

I(h, k) =I₁+I₂ = log{α}+ log{−α}

h + 1

hV₁(^h_k, α) + 1

kV₂({_h^k}, β) +log(k/h)

k +O

1 k + 1

h

. Comparing this with (2.5) we obtain the claimed identity, since for 0< x <1,y >0 we have

logx+ log(1−x)−log⁻(xy)−log⁻((1−x)y) =−γlogy+O(1),

whereγ = 2 if 0< y≤1 and 0< γ ≤1 otherwise.

2.3. Bound for V₁ in terms of the continued fraction expansion. We now iterate the relation (2.4) in order to obtain a bound for individual values ofV1.

Corollary 2. Let `, h, k∈Nwithh < kand(h, k) = 1. Then, uniformly forα∈ ¹_kZ, we have

V1

h k, α

≤

r−1

X

m=0

log(v_m+1/v_m) vm

+O(1),

V1

h k, α

≤ 1 k

r

X

m=1

vmlog(vm/vm−1) +O(1), where v_m is them-th partial quotient of ^h_k.

Proof. By Proposition 1, recalling the notation β :={^k_h}⁻¹{_h^`} ifk -` and β := 0 otherwise, we have

kV1

h k, `

k

≤δh6=1

hV2

nk h

o , β

+klog k

h

+O(k) ifh≤k, (2.7)

hV₂

k h, β

≤

kV₁

h k,`

k

+hlog h

k

+O(h) ifh≥k, h6= 1.

(2.8)

Now, leth/k= [0;b1, . . . , br] be the continued fraction expansion of h/k, withbr 6= 1 ifr >1.

Also, leth^∗ ∈[1, k] be such that h^∗ ≡(−1)^r+1h (modk) (in particular h^∗/k= [0, b_r, . . . , b₁]).

The Euclid algorithm onh^∗ and kcan be written as (see [Khi63]) v_r =k, vr−1=h^∗,

v`+1 =b`+1v`+v`−1, `= 0, . . . r.

Then, alternating the use of (2.7)-(2.8) and the reduction modulo the denominator inV₁, we obtain

(2.9) k V₁

h k, α

≤

r

X

m=1

v_mlog(v_m/vm−1) +O(v_m)

=

r

X

m=1

v_mlog(v_m/vm−1) +O(k), as desired, where the last step follows sincevn−2 ≤v_n/2 for alln.

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Indicating withu_m/v_mandu⁰_m/v_m⁰ them-th convergents ofh/kandh^∗/krespectively (with v−1=v⁰₋₁ = 0), one hask=vsv_r−s⁰ +vs−1v_r−s−1⁰ for all 0≤s≤r (see [Hei69, p. 91-92]). In particular,k/2≤v_sv⁰_r−s ≤k. Thus, by (2.9) we have

k V1

h k, α

≤

r

X

m=1

v_m⁰ log(v_m⁰ /v_m−1⁰ ) +O(k)≤k

r−1

X

m=0

log(vm+1/vm) vm

+O(k).

3. Proof of Theorem 1

We will deduce Theorem 1 from Corollary 2 and the following lemma, which shows that S_f ^h_k

is, up to a small error, a linear combination of V₁(^h_k,^`_k^j) for various values of`_j. Lemma 3. Let h ∈ Z, k ∈ N with (h, k) = 1. Let f : R → R be a 1-periodic function which is piecewise C¹, and continuous everywhere except possibly at d points ^`_k¹, . . . ,^`_k^d, with 0 ≤ `₁ < · · · < `_d < k, and assume f is left and right differentiable at such points. Also, assume 2f(^`_kⁱ) =f(^`_kⁱ+) +f(^`_kⁱ−),where f(x±) := lim_ε→0+f(x±ε). Then,

S_f h

k

= 1 π

d

X

j=1

V₁ h

k,`_i k

f`_j

k +

−f`_j k −

+O

kf⁰k^1/2₂ .

Proof. We have the Fourier expansion f(x) =X

n∈Z

f(n) e(−nx),ˆ x∈R where, for n6= 0,

f(n) =ˆ

d

X

j=1

f(^`_k^j−)−f(^`_k^j+)

2πin e(n`_j/k)−fˆ⁰(n) 2πin, and fb⁰(n) :=R1

0 f⁰(y) e(ny)dy. Note that, by the Bessel inequality,

(3.1) X

n6=0

fb⁰(n) n

kf⁰k₂.

Now, we have

k−1

X

m=1

cot

πmh k

e

−nm k

=−2ik nh

k

and thus,

k−1

X

m=1

fm k

cot πmh

k

=−2ik X

n∈Z6=0

nh k

d

X

j=1

f(^`_k^j−)−f(^`_k^j+)

2πin e(n`_j/k)−fb⁰(n) 2πin

.

Grouping this with the definition (2.3) and the bound (3.1) yields our claim.

Proof of Theorem 1. At the cost of committing an error of sizeO(dD0) inSf h k

, we modify f so that all of its d points of discontinuity are at rationals of the form _k^` with 2f ^`_k

= f _k^`+

+f _k^`−

. Theorem 1 then follows by Proposition 1 and Lemma 3.

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Particular case: partial cotangent sums. For the partial cotangent sums, alluded to in the introduction (Figure 1), since there is only one point of discontinuity other than 0, we indeed obtain a (non-exact) reciprocity relation in the following form.

Corollary 4. Let h ∈ Z, k ∈ N with (h, k) = 1, 1 ≤ h < k, and let 0 < ` < k. Let k1 ≡k(mod h),`⁰ ≡`(modh)with1≤k1, `⁰ ≤h and let`1≡`⁰ (modk1), h1 ≡h (modk1) with0≤h₁< k₁, 1≤`₁ ≤k₁. Then,

C_`(h/k)−C_`₁(h1/k1) = 1 π

γ_h,k,`−1 logk

h +O h

k₁

. with0≤γh,k,`≤1.

Proof. By Lemma 3 we haveC_`(h/k) = ¹_π(V1(^h_k,_k^`)−V(^h_k,0)) +O(1). The result then follows, since applying Proposition 1 to (^h_k,_k^`) and to (_k^h

1,_k^`⁰

1)≡(^h_k¹

1,_k^`¹

1) (mod 1) we obtain V₁

h k, `

k

−V₁ h

k1

, `⁰ k1

=

γ_h,k,`−h k

logk h + h

k1

logk₁ h +O

h k1

.

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